Bi-elliptic transfer

Example

Let $\mu > 0$, let $0 < r_1 < r_2$ be the radii of two coplanar circular orbits about the same centre, and choose $r_b \ge r_2$. The bi-elliptic transfer uses two half-ellipses, both Kepler orbits. A first tangential burn at $r_1$ puts the craft on an ellipse with periapsis $r_1$ and apoapsis $r_b$, of semi-major axis $a_1 = (r_1 + r_b)/2$. At $r_b$ a second burn raises the periapsis to $r_2$, giving an ellipse of semi-major axis $a_2 = (r_2 + r_b)/2$. At $r_2$ a third, braking, burn puts the craft on the outer circle. By the vis-viva equation the three changes of speed are $$\Delta v_1 = \sqrt{\frac{2\mu}{r_1} - \frac{\mu}{a_1}} - \sqrt{\frac{\mu}{r_1}}, \quad \Delta v_2 = \sqrt{\frac{2\mu}{r_b} - \frac{\mu}{a_2}} - \sqrt{\frac{2\mu}{r_b} - \frac{\mu}{a_1}}, \quad \Delta v_3 = \sqrt{\frac{2\mu}{r_2} - \frac{\mu}{a_2}} - \sqrt{\frac{\mu}{r_2}}.$$

Check with numbers: take $\mu = 1$, $r_1 = 1$, $r_2 = 20$ and let $r_b \to \infty$. Then $\Delta v_1 \to \sqrt{2} - 1$, $\Delta v_2 \to 0$ and $\Delta v_3 \to (\sqrt{2} - 1)/\sqrt{20}$, a total of about $0.507$. The Hohmann transfer between the same circles needs $\Delta v_1 + \Delta v_2 \approx 0.380 + 0.155 = 0.535$. So here the bi-elliptic transfer is cheaper.